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Applications of Gauss' Law - two hollow spheres

  • Thread starter knowlewj01
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  • #1
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Homework Statement



A hollow sphere of radius [itex]r_1[/itex] is placed at the centre of a larger hollow sphere of radius [itex]r_2[/itex].
Both spheres have a uniformly distributed total charge of +q
Find the preassure [itex]p(r_2 , q)[/itex] which acts on the outer sphere.

Homework Equations



[itex]\oint\textbf{E.n}dS = 4\pi k Q[/itex]

[itex]p = \frac{dF}{dA}[/itex]

The Attempt at a Solution



Inside the larger sphere, there is no contribution to electric field from the large sphere, so just consider the electric field at [itex]r_2[/itex] due to the inner sphere:

[itex]E(4\pi r_2^2) = 4\pi k q[/itex]

[itex]E = \frac{kq}{r_2^2}[/itex]

since [itex]F = qE[/itex]

the force acting at [itex]r_2[/itex]:

[itex]F(r_2) = \frac{kq^2}{r_2^2}[/itex]

[itex]p = \frac{F}{A}[/itex]

[itex]p = \frac{kq^2}{4\pi r_2^4}[/itex]

does this look right?
 
Last edited:

Answers and Replies

  • #2
110
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Just had a thought, at the point when calutating force, replace k with [itex]\frac{1}{4\pi \epsilon_0}[/itex]

[itex]F(r_2) = \frac{q^2}{4\pi r_2^2 \epsilon_0}[/itex]

the expression for the total area, A, of the outer sphere is now on the denominator:

[itex] A = 4\pi r_2^2[/itex]

it follows that:

[itex] F = \frac{q^2}{A\epsilon_0}[/itex]

now differentiate F with respect to A:

[itex]\frac{dF}{dA} = -\frac{q^2}{A^2 \epsilon_0} = p[/itex]

[itex] p = \frac{q^2}{16\pi^2 r_2^4 \epsilon_0}[/itex]

gthis is the same answer i got before but i guess it's a more thorough way of doing it.
 
Last edited:
  • #3
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i took the center sphere to act as a point charge, and calculated E then F then P

got exactly the same. so looks good?
 

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